Question: Let $a(x)=5x^{10}-9x^8-9x^5+7x$, and $b(x)=x^5$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's rewrite the fraction to cancel common factors: $ \begin{aligned} \dfrac{a(x)}{b(x)}=\dfrac{5x^{10}-9x^8-9x^5+7x}{x^5}&=\dfrac{5 {x^{10}}-9 {x^8}-9 {x^5}}{ {x^5}}+\dfrac{7x}{x^5}\\\\ &={5x^5-9x^3-9}+\dfrac{{7x}}{x^5}\\\\ &={q(x)} + \dfrac{{r(x)}}{b(x)}\end{aligned}$ Since the degree of ${7x}$ is less than the degree of $x^5$, it follows that ${r(x)}={7x}$, and ${q(x)}={5x^{5}-9x^3-9}$. To conclude, $q(x)=5x^{5}-9x^3-9$ $r(x)=7x$ [Is there another way of doing this?]